Internal problem ID [1503]
Internal file name [OUTPUT/1504_Sunday_June_05_2022_02_19_54_AM_73131463/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 9 Introduction to Linear Higher Order Equations. Section 9.3. Undetermined
Coefficients for Higher Order Equations. Page 495
Problem number: section 9.3, problem 6.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_3rd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime }+y^{\prime \prime }-2 y={\mathrm e}^{x} \left (15 x^{2}+34 x +14\right )} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime }+y^{\prime \prime }-2 y = 0 \] The characteristic equation is \[ \lambda ^{3}+\lambda ^{2}-2 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1-i\\ \lambda _3 &= -1+i \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1-i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{x} \\ y_2 &= {\mathrm e}^{\left (-1-i\right ) x} \\ y_3 &= {\mathrm e}^{\left (-1+i\right ) x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+y^{\prime \prime }-2 y = {\mathrm e}^{x} \left (15 x^{2}+34 x +14\right ) \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ {\mathrm e}^{x} \left (15 x^{2}+34 x +14\right ) \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{{\mathrm e}^{x} x, {\mathrm e}^{x} x^{2}, {\mathrm e}^{x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{{\mathrm e}^{x}, {\mathrm e}^{\left (-1-i\right ) x}, {\mathrm e}^{\left (-1+i\right ) x}\} \] Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{{\mathrm e}^{x} x, {\mathrm e}^{x} x^{2}, {\mathrm e}^{x} x^{3}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{1} {\mathrm e}^{x} x +A_{2} {\mathrm e}^{x} x^{2}+A_{3} {\mathrm e}^{x} x^{3} \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ 5 A_{1} {\mathrm e}^{x}+10 A_{2} {\mathrm e}^{x} x +8 A_{2} {\mathrm e}^{x}+15 A_{3} {\mathrm e}^{x} x^{2}+24 A_{3} {\mathrm e}^{x} x +6 A_{3} {\mathrm e}^{x} = {\mathrm e}^{x} \left (15 x^{2}+34 x +14\right ) \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 0, A_{2} = 1, A_{3} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = {\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x^{3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1-i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3}\right ) + \left ({\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x^{3}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1-i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x^{3} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{x}+{\mathrm e}^{\left (-1-i\right ) x} c_{2} +{\mathrm e}^{\left (-1+i\right ) x} c_{3} +{\mathrm e}^{x} x^{2}+{\mathrm e}^{x} x^{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+y^{\prime \prime }-2 y={\mathrm e}^{x} \left (15 x^{2}+34 x +14\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=15 \,{\mathrm e}^{x} x^{2}+34 \,{\mathrm e}^{x} x +14 \,{\mathrm e}^{x}-y_{3}\left (x \right )+2 y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=15 \,{\mathrm e}^{x} x^{2}+34 \,{\mathrm e}^{x} x +14 \,{\mathrm e}^{x}-y_{3}\left (x \right )+2 y_{1}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & -1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 15 \,{\mathrm e}^{x} x^{2}+34 \,{\mathrm e}^{x} x +14 \,{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 15 \,{\mathrm e}^{x} x^{2}+34 \,{\mathrm e}^{x} x +14 \,{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 0 & -1 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ], \left [-1-\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [-1+\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{2} \\ -\frac {1}{2}-\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-1-\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-\mathrm {I}\right ) x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{-x}\cdot \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \\ -\frac {1}{2}+\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{2} \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \left (-\frac {1}{2}+\frac {\mathrm {I}}{2}\right ) \left (\cos \left (x \right )-\mathrm {I} \sin \left (x \right )\right ) \\ \cos \left (x \right )-\mathrm {I} \sin \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\sin \left (x \right )}{2} \\ -\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{-x}\cdot \left [\begin {array}{c} -\frac {\cos \left (x \right )}{2} \\ \frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2} \\ -\sin \left (x \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{x} & -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \cos \left (x \right ) & -{\mathrm e}^{-x} \sin \left (x \right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{x} & -\frac {{\mathrm e}^{-x} \sin \left (x \right )}{2} & -\frac {{\mathrm e}^{-x} \cos \left (x \right )}{2} \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \left (-\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) & {\mathrm e}^{-x} \left (\frac {\cos \left (x \right )}{2}+\frac {\sin \left (x \right )}{2}\right ) \\ {\mathrm e}^{x} & {\mathrm e}^{-x} \cos \left (x \right ) & -{\mathrm e}^{-x} \sin \left (x \right ) \end {array}\right ]\cdot \left (\left [\begin {array}{ccc} 1 & 0 & -\frac {1}{2} \\ 1 & -\frac {1}{2} & \frac {1}{2} \\ 1 & 1 & 0 \end {array}\right ]\right )^{-\mathrm {1}} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \frac {\left (3 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-2 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-\cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x}}{5} \\ \frac {\left (-4 \sin \left (x \right )-2 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (3 \cos \left (x \right )+\sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-\cos \left (x \right )+3 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x}}{5} \\ \frac {\left (-2 \cos \left (x \right )+6 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (-4 \sin \left (x \right )-2 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {2 \,{\mathrm e}^{x}}{5} & \frac {\left (4 \cos \left (x \right )-2 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x}}{5} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} \frac {\left (4 \sin \left (x \right )+2 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x} \left (5 x^{3}+5 x^{2}-2\right )}{5} \\ \frac {\left (2 \cos \left (x \right )-6 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\left (2 x +4 x^{2}-\frac {2}{5}+x^{3}\right ) {\mathrm e}^{x} \\ \frac {\left (-8 \cos \left (x \right )+4 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\left (10 x +7 x^{2}+\frac {8}{5}+x^{3}\right ) {\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\left [\begin {array}{c} \frac {\left (4 \sin \left (x \right )+2 \cos \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{x} \left (5 x^{3}+5 x^{2}-2\right )}{5} \\ \frac {\left (2 \cos \left (x \right )-6 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\left (2 x +4 x^{2}-\frac {2}{5}+x^{3}\right ) {\mathrm e}^{x} \\ \frac {\left (-8 \cos \left (x \right )+4 \sin \left (x \right )\right ) {\mathrm e}^{-x}}{5}+\left (10 x +7 x^{2}+\frac {8}{5}+x^{3}\right ) {\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (-5 c_{3} +4\right ) \cos \left (x \right )+\left (-5 c_{2} +8\right ) \sin \left (x \right )\right ) {\mathrm e}^{-x}}{10}+\left (c_{1} +x^{3}+x^{2}-\frac {2}{5}\right ) {\mathrm e}^{x} \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] trying high order linear exact nonhomogeneous trying differential order: 3; missing the dependent variable checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 30
dsolve(diff(y(x),x$3)+diff(y(x),x$2)-2*y(x)=exp(x)*(14+34*x+15*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{2} \cos \left (x \right )+c_{3} \sin \left (x \right )\right ) {\mathrm e}^{-x}+{\mathrm e}^{x} \left (x^{3}+x^{2}+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.159 (sec). Leaf size: 49
DSolve[y'''[x]+y''[x]-2*y[x]==Exp[x]*(14+34*x+15*x^2),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{5} e^{-x} \left (e^{2 x} \left (5 x^3+5 x^2-2+5 c_3\right )+5 c_2 \cos (x)+5 c_1 \sin (x)\right ) \]